[SI-LIST] Re: Is a copper plane not tied to any net a reference plane?
- From: Larry Smith <Larry.Smith@xxxxxxx>
- To: ericsilist@xxxxxxxxx
- Date: Tue, 04 May 2004 21:27:32 -0700
Who is right? You both are. The return current is on the plane next
to the trace for the first few nSec then moves to the Gnd plane in the
normal stackup.
Here is my reasoning. Initially, all of the return current is on the
nearby plane. Assume that the trace dimension is chosen to make a 50
Ohm transmission line with air dielectric. The nearby plane controls
the impedance of the trace even though the dielectric constant of air
is 1 compared to about 4 for FR4. The top plane is 20 times closer to
the trace than the Gnd plane and most of the fields (electric and
magnetic) are found between the trace and sheet of copper. Current
travels where the magnetic field lines appear to terminate in the
copper.
Where does the current come from? It comes from charge on the
surrounding copper plane. A radial wave goes out from point close to
the start of the trace and proceeds at a velocity that is somewhat
greater than half the speed of light as determined by the square roots
of the relative dielectric constants of the FR4 and air. During the 2
nSec rise time, the radial wave travels to at least a 12 inch radius
circle. Assuming a positive voltage step input, charge accumulates on
the plane near the trace and is depleted from the Gnd plane 105 mils
below, thus completing the return path for the long via with
displacement current. If my calculations are correct, the capacitance
of the parallel plate "reference plane" capacitor is 7.5 pF/square
inch giving 3.4 nF of capacitance for the 12 inch radius circle during
the rise time. The frequency content of the rise time is .35/2nSec =
180MHz. At 180 MHz, 3.4 nF has .27 Ohms of impedance, much less than
the 50 Ohm transmission line. Therefor, most of the voltage drop and
energy of the transient will go into the 50 Ohm transmission line and
only a small percentage will go into making the planes bounce.
This continues until the rising edge reaches the end of the trace.
Assume that the trace is 30 inches long and then goes back down into
the normal stackup. Five nSec after the event started, the wavefront
reaches the end of the trace and return current is flowing in the
opposite direction on the nearby reference plane for the entire length
of the trace. That current has to come from somewhere and starts
depleting the charge from the reference plane near the end of the
trace. Just as before, the floating plane and Gnd plane begin to
bounce setting up another radial wave centered at the end of the
trace, but in the opposite polarity as before. This is the beginning
of the end for the net current on the upper plane. Charge is being
depleted from the plane on one end of the trace and accumulating on
the other end causing a voltage difference. Most people have seen an
animated simulation from Sigrity or other software that shows how the
power planes bounce during such an event. The potentials on the
planes are such that eddy currents will start up and soon circular
currents are traveling under the trace and completing the loop out on
the broad plane somewhere. It has to, otherwise a lot of charge would
build up on one end of the board. And an equal and opposite eddy
current begins flowing on the Gnd plane 105 mils below the floating
plane. So, within about 2 time of flights of the trace, return
current is flowing both under the trace on the floating plane and in
big circular patterns in the Gnd plane.
If we wait long enough without disturbing the signal on the trace, the
return current under the trace and the eddy currents in the floating
plane will eventually die out. Certainly at DC, there is no current
on the floating plane. Assuming about 10 nH/inch for the trace and 1
mOhm/square for the plane, I calculate that this will happen about 20
uSec after the event (20,000 rise times later).
regards,
Larry Smith
Sun Microsystems
eric steimle wrote:
>
> I?m trying to settle an argument without saying which
> side I?m on, and I was hoping someone could give a
> quick and easy example to prove this.
>
> You have a four layer board that looked like this
>
> SIG1
> GND
> VCC
> SIG2
>
> And the maximum rise time on the board was about 2ns
> (say a 100MHz clock), then if I ran a trace from one
> IC to another on SIG1 the return current would flow
> along the GND plane (assuming no splits in the plane
> etc.)
>
> What if I covered SIG1 with a plane of copper say
> hovering 5 mil above SIG1, and then made the distance
> between SIG1 and GND 100mil so..
>
> Plane sheet of Copper
> 5 mil of air
> SIG1
> 100 mil FR4
> GND
> VCC
> SIG2
>
> Now that plane sheet is not GND and it has no
> association to any net it is just a continuous sheet
> of copper about the size of the board. One of us
> argues that the return current will continue to flow
> along the GND plane as it did before. The other
> argues that the return current will instead flow
> mostly along that sheet of copper unitl it jumps back
> to the GND plane as it gets closer to the IC. Any
> help in this would be much appreciated.
>
> Thanks
>
>
>
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